integral from 1 to-1 of (x-1)(3x+2)

Lösung

\int_{1}^{- 1} (x - 1) (3 x + 2) d x

2

Schritte zur Lösung

\int_{1}^{- 1} (x - 1) (3 x + 2) d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{1} (x - 1) (3 x + 2) d x

Multipliziere aus

(x - 1) (3 x + 2) : 3 x^{2} - x - 2

= - \int_{- 1}^{1} 3 x^{2} - x - 2 d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int_{- 1}^{1} 3 x^{2} d x - \int_{- 1}^{1} x d x - \int_{- 1}^{1} 2 d x)

\int_{- 1}^{1} 3 x^{2} d x = 2

\int_{- 1}^{1} x d x = 0

\int_{- 1}^{1} 2 d x = 4

= - (2 - 0 - 4)

Vereinfache

= 2

\int_{10}^{13} 2 x d x

d er i v a t i v e y = 4 x - 3 x^{2}

x \to \infty lim (\frac{4 ^{x}}{3 ^{x - 1}})

\frac{\partial}{\partial x} (\frac{y}{y ^{2} + x ^{2}})

\int_{2}^{8} (10 - x) d x